Let $f_1:[a,b]\rightarrow \mathbb{R}$ be a Riemann integrable function. Define the sequence of functions $f_n:[a,b] \rightarrow \mathbb{R}$ by
$f_{n+1}(x)=\int_a^x f_n(t)dt,$
for each $n\ge 1$ and each $x\in [a,b]$. Prove that the sequence of functions
$g_n(x)=\sum\limits_{k=1}^n f_k(x)$
converges uniformly on [a,b].
I try to use Ascoli Theorem to prove this problem by considering $|g_n(x)-gn(y)|\le\sum\limits_{k=1}^n |f_k(x)-f_k(y)|$. However, $|f_k(x)-f_k(y)|\le M |b-a|^{k-2} |x-y|, $ where $M>0$ is a constant achieving by the condition that $f_1$ is an integrable function. However, this estimation depends on $k$, so I cannot apply Ascoli-Azela Theorem.
Thank you in advance for your help.
Thanks to @RRL comment, I rewrite the solution as follows.
Using the following theorem to prove
"Suppose $\{f_n\}$ is a sequence of functions defined on E, and suppose $|f_n(x)|\le M_n$ ($x\in E$, $n=1,\ 2,\ 3,\ldots$). Then $\sum f_n$ converges uniformly on $E$ if $\sum M_n$ converges."
Since $f_1$ is a Riemann integrable function, there is a constant $M>0$ such that $|f_1(x)|\le M$ for all $x\in [a,b]$. From the assumption $f_{n+1}(x)=\int_a^x f_n(t)dt,$ we easily get the following inequalities: $$f_2(x)\le M(b-a),$$ $$f_3(x)\le \frac{1}{2!}M(b-a)^2,$$ $$\ldots$$ $$f_{n+1}(x)\le \frac{1}{n!}M(b-a)^n,$$
We have $M+M(b-a)+\frac{1}{2!}M(b-a)^2+ \ldots +M \frac{1}{n!}(b-a)^n+ \ldots = Me^{b-a}$. Applying the above theorem we have $g_n=\sum\limits_{k=1}^n f_k(x)$ converges uniformly on $[a,b]$.